How do you evaluate the limit tan(3x)/(3tan2x) as x approaches 0?

Mar 6, 2017

$L {t}_{x \to 0} \frac{\tan \left(3 x\right)}{3 \tan 2 x} = \frac{1}{2}$

Explanation:

Let us first find $L {t}_{x \to 0} \tan \frac{x}{x}$

$L {t}_{x \to 0} \tan \frac{x}{x} = L {t}_{x \to 0} \frac{\sin x}{x \cos x}$

= $L {t}_{x \to 0} \frac{\sin x}{x} \times L {t}_{x \to 0} \frac{1}{\cos} x$

= $1 \times 1 = 1$

Hence $L {t}_{x \to 0} \frac{\tan \left(3 x\right)}{3 \tan 2 x}$

= $L {t}_{x \to 0} \frac{\frac{\tan \left(3 x\right)}{3 x}}{3 \frac{\tan 2 x}{2 x}} \times \frac{3 x}{2 x}$

= $\frac{L {t}_{3 x \to 0} \left(\frac{\tan 3 x}{3 x}\right)}{3 L {t}_{2 x \to 0} \left(\frac{\tan 2 x}{2 x}\right)} \times \frac{3}{2}$

= $\frac{1}{3 \times 1} \times \frac{3}{2} = \frac{1}{2}$

Mar 6, 2017

${\lim}_{x \rightarrow 0} \tan \frac{3 x}{3 \tan \left(2 x\right)} = \frac{1}{2}$

Explanation:

Use $\tan \left(a + b\right) = \frac{\tan \left(a\right) + \tan \left(b\right)}{1 - \tan \left(a\right) \tan \left(b\right)}$ to rewrite $\tan \left(3 x\right)$ as $\tan \left(2 x + x\right)$:

${\lim}_{x \rightarrow 0} \tan \frac{3 x}{3 \tan \left(2 x\right)} = {\lim}_{x \rightarrow 0} \frac{\frac{\tan \left(2 x\right) + \tan \left(x\right)}{1 - \tan \left(2 x\right) \tan \left(x\right)}}{3 \tan \left(2 x\right)}$

$= {\lim}_{x \rightarrow 0} \frac{\tan \left(2 x\right) + \tan \left(x\right)}{3 \tan \left(2 x\right) \left(1 - \tan \left(x\right) \tan \left(2 x\right)\right)}$

Now using $\tan \left(2 x\right) = \frac{2 \tan \left(x\right)}{1 - {\tan}^{2} \left(x\right)}$:

$= {\lim}_{x \rightarrow 0} \frac{\frac{2 \tan \left(x\right)}{1 - {\tan}^{2} \left(x\right)} + \tan \left(x\right)}{3 \frac{2 \tan \left(x\right)}{1 - {\tan}^{2} \left(x\right)} \left(1 - \tan \left(x\right) \frac{2 \tan \left(x\right)}{1 - {\tan}^{2} \left(x\right)}\right)}$

Multiplying through by $1 - {\tan}^{2} \left(x\right)$:

=lim_(xrarr0)(2tan(x)+tan(x)(1-tan^2(x)))/(6tan(x)(1-(2tan^2(x))/(1-tan^2(x)))

Factoring $\tan \left(x\right)$ then cancelling:

=lim_(xrarr0)(tan(x)(2+1-tan^2(x)))/(6tan(x)(1-(2tan^2(x))/(1-tan^2(x)))

=lim_(xrarr0)(2+1-tan^2(x))/(6(1-(2tan^2(x))/(1-tan^2(x)))

Now we can plug in $x = 0$ and evaluate the limit without a problem:

$= \frac{2 + 1 - 0}{6 \left(1 - \frac{2 \left(0\right)}{1 - 0}\right)} = \frac{3}{6} = \frac{1}{2}$